Can we solve it with the rotational KE and its relation to PE?TSny said:Welcome to PF!
##\sum \tau = I \alpha## only applies to a rigid body rotating about a fixed axis. Here, the Earth is not acting as a rigid body.
Can you relate this problem to any other examples you've been studying lately where a rotating system changes its mass distribution while no external torque acts?
Rather than energy, there's a much more appropriate physical quantity that can be used. Have you read or discussed in class what goes on when an ice skater pulls in her arms in order to spin faster?ht9000 said:Can we solve it with the rotational KE and its relation to PE?
Yes. (Distinguish between the two ω's.)ht9000 said:Oh yeah! Conservation of angular momentum! Which is L = Iω. And it will be the same at its original size and after it grows a little bit.
(2/5)mr^2 ω = (2/5)m(r+.03)^2 ω, m's cancel out and I solve for ω?
Right now what I have is:TSny said:OK. But since you are asked to find the % change in ω, you could first work out an expression for the % change and simplify the expression before substituting numbers. That way, you might not even need to know the initial rate of spin.
You divide it by the original quantity and multiply by 100TSny said:Yes. How to you find the % change in a quantity?
So it will be the change in the period divided by the original period.TSny said:
Is this right?TSny said:It might be helpful to note that ##\frac{T_f - T_i}{T_i} = \frac{T_f}{T_i}-1##.
No. I would go back to the angular momentum relation and express it in terms of T rather than ω. Then you can easily find an expression for Tf/Ti.ht9000 said:to get that expression for the change in T, would I just have to take 2π divided by the expression I have now?
ω in terms of T is (Δθ)/T right? I substituted that in place of ω and when I solved for it, it gave meTSny said:No. I would go back to the angular momentum relation and express it in terms of T rather than ω. Then you can easily find an expression for Tf/Ti.
Yes, with Δθ = 2##\pi##.ht9000 said:ω in terms of T is (Δθ)/T right?
Yes, that's it.I substituted that in place of ω and when I solved for it, it gave me
((r_f)^2 / (r_i)^2) - 1
Is that the expression for % change in the period?
Awesome, thanks a lot for the helpTSny said:Yes, that's it.
Rotational period, also known as rotational speed or angular velocity, is the measure of how quickly an object rotates around its axis.
No, rotational period is not affected by an object's mass. It is solely determined by the object's size and shape.
As the volume of an object increases, its rotational period decreases. This is because the object's mass is spread out over a larger area, resulting in a larger moment of inertia and slower rotation.
A frisbee and a basketball have the same mass, but the frisbee has a larger rotational period due to its larger volume and lower moment of inertia. Similarly, a spinning top and a coin have different rotational periods despite having the same mass.
Objects with longer rotational periods tend to be more stable, as they have a lower angular velocity and are less likely to topple over. This is why it is easier to balance a larger spinning top than a smaller one.